# Problem of a function

• Jan 26th 2007, 09:39 AM
Problem of a function
Suppose that $\displaystyle f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous, bounded, strictly increasing function.

Questions:

(a) Show that there is a point $\displaystyle a_{1}\epsilon\mathbb{R}$ such that $\displaystyle f(a_{1})>a_{1}$
(b) For each $\displaystyle n\epsilon\mathbb{N}$, define $\displaystyle a_{n+1}=f(a_{n})$. Explain why $\displaystyle a_{2}>a_{1}$. Then explain why $\displaystyle a_{n+1}>a_{n}$ for all $\displaystyle n\epsilon\mathbb{N}$
(c) Explain why the sequence $\displaystyle {a_{n}}$ is bounded above.
(d) Explain why the sequence $\displaystyle {a_{n}}$ converges to some number, L.
(e) Explain why f(L) = L.

My work so far:

(a) f is bounded and strictly increasing, therefore f'(x) > 0, so there is a point $\displaystyle x\epsilon\mathbb{R}$ such that x < f(x). I know this reason is not enough, still working on the proper proof.

(b) Now $\displaystyle a_{2}=a_{1+1}=f(a_{1})>a_{1}$ as defined by part (a).
Let $\displaystyle S = {n \epsilon \mathbb{N}:a_{n+1}>a_{n}}$
$\displaystyle 1 \epsilon S$ as $\displaystyle a_{1+1}=a_{2} > a_{1}$
let $\displaystyle k \epsilon S$, then $\displaystyle a_{k+1} > a_{k}$
Now $\displaystyle a_{k+1+1} = f(a_{k+1}) > a_{k+1}$, thus proved $\displaystyle k+1 \epsilon S$
Therefore S is inductive, and proved $\displaystyle a_{n+1}>a_{n}$

(c) Because f is continuous and bounded.

(d) Because f is monotonic, implies $\displaystyle {a_{n}}$ is monotonic, thus $\displaystyle {a_{n}}$ is bounded. All bounded monotonic sequence converges, thus the sequence converges to L.

(e) $\displaystyle {a_{n}}$converges to L, then $\displaystyle f({a_{n}})$ converges to L because f is continuous.

I know I made some mistakes up there, please check, thank you.

KK
• Jan 26th 2007, 09:55 AM
ThePerfectHacker
Quote:

(a) Show that there is a point $\displaystyle a_{1}\epsilon\mathbb{R}$ such that $\displaystyle f(a_{1})>a_{1}$

Let us assume that $\displaystyle f(x)\geq x$ for all $\displaystyle x\in \mathbb{R}$. Then $\displaystyle f(x)$ is not bounded because the function $\displaystyle x$ is not bounded. Which is a contradiction.

FOR ALL $\displaystyle x\in \mathbb{R}$ we have $\displaystyle f(x)\geq x$.

The negation of that (which must be because by contradiction) is,
FOR SOME $\displaystyle x\in \mathbb{R}$ we have $\displaystyle f(x)<x$.
Because the negation of a universal quantifier is an existencial quantifier. And the negation of $\displaystyle \geq$ is $\displaystyle <$.
• Jan 26th 2007, 10:04 AM
ThePerfectHacker
Quote:

(a) f is bounded and strictly increasing, therefore f'(x) > 0, so there is a point $\displaystyle x\epsilon\mathbb{R}$ such that x < f(x). I know this reason is not enough, still working on the proper proof.

The problem with that is we do not know whether $\displaystyle f$ is differenciable or not.

Quote:

(b) Now $\displaystyle a_{2}=a_{1+1}=f(a_{1})>a_{1}$ as defined by part (a).
Let $\displaystyle S = {n \epsilon \mathbb{N}:a_{n+1}>a_{n}}$
$\displaystyle 1 \epsilon S$ as $\displaystyle a_{1+1}=a_{2} > a_{1}$
let $\displaystyle k \epsilon S$, then $\displaystyle a_{k+1} > a_{k}$
Now $\displaystyle a_{k+1+1} = f(a_{k+1}) > a_{k+1}$, thus proved $\displaystyle k+1 \epsilon S$
Therefore S is inductive, and proved $\displaystyle a_{n+1}>a_{n}$
I got the same thing. It is inductive because $\displaystyle a_{n}>a_{n-1}$ then $\displaystyle a_{n+1}=f(a_n)>f(a_{n-1})=a_n$ because it is strictly increasing.
• Jan 26th 2007, 10:07 AM
I'm sorry, I don't really understand what you were doing. Did you proved f(x) < x by contradiction? But I'm trying to prove f(x) > x.

What is negation?

Sorry about my lack of understanding.

KK
• Jan 26th 2007, 10:10 AM
ThePerfectHacker
Quote:

I'm sorry, I don't really understand what you were doing. Did you proved f(x) < x by contradiction? But I'm trying to prove f(x) > x.

What is negation?

Sorry about my lack of understanding.

KK

Sorry I proved a statement that you are not looking for.
• Jan 26th 2007, 10:12 AM
Quote:

Originally Posted by ThePerfectHacker
The real number $\displaystyle f(a_1)$ is an upper bound.
Because,
$\displaystyle f(a_1)>a_1>a_2>a_3>...$.

Isn't $\displaystyle a_{n+1}>a_n$? So shouldn't $\displaystyle a_1<a_2<a_3...$?

So would you say f is bounded above by $\displaystyle f(a_n)$?

Thanks.

KK
• Jan 26th 2007, 10:15 AM
ThePerfectHacker
Quote:

Isn't $\displaystyle a_{n+1}>a_n$? So shouldn't $\displaystyle a_1<a_2<a_3...$?
So would you say f is bounded above by $\displaystyle f(a_n)$?
No you cannot say $\displaystyle f(a_n)$ an an upper bounded because $\displaystyle a_n$ is not a real number.